Friday, December 6, 2013

Eigenvectors of recurrence equations

I have watched the Gilber Strang's great course. Lec 22 | MIT 18.06 Linear Algebra, explains everything but what the eigenvectors are. After some struggling I have figured out that companion matrix turns eighenvector [a b c d] into [λa λb λc λd]. Herein, λ is vector's the eigenvalue. Now, what are the [a, b, c, etc]? They are [λ³ λ² λ 1] because a element of the input vector turns into λa at the next iteration! Next time, λa turns into λ²a. This way, you must conclude that the train of values is the vector [λ³ λ² λ 1]. If we recall that it corresponds to [y₃  y₂  y₁  y₀], we'll understand that we start in y₀, next state variable y₁ = λy₀, y₂ = λ²y₀ and progress so on. But, we can factor the y₀ out of the eigenvector, [y₃  y₂  y₁  y₀] = y₀ [λ³ λ² λ 1] and, because all eigenvectors are identical up the scale factor, we can safely eliminate the y₀, yielding the eigenvector of  [λ³ λ² λ 1]. It is perfect that this vector is a geometric series (aka exponential function), which is an eigenfunction of any differential and difference equation.

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